Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Does anybody know an easy explanation of the proof of Artin's vanishing theorem (that the etale cohomology of an affine variety of dimension $n$ over an algebraically closed field vanishes in degrees $>n$, or of any other version of this statement)? I have found some proofs; all of them are step by step, and it is not clear to me which of these steps are the most important ones. So, what is the central idea here?

share|improve this question
add comment

1 Answer

up vote 9 down vote accepted

I was curious myself after learning this result sometime ago from Lazarsfeld's book on positivity (he calls it the Artin-Grothendieck theorem). The corresponding statement for smooth varieties over the complex numbers and singular cohomology (theorem of Andreotti-Frankel) follows from the fact that Morse theory shows that the variety is homotopy equivalent to a CW-complex with no cells in dimensions $>n$.

The etale cohomology counterpart works more generally for constructible sheaves. This is probably not very helpful, but here is a sketch of the argument from Lazarsfeld (he uses constructible sheaves in the complex topology, but it should adapt to etale sheaves):

  1. Reduce to the affine space by choosing a finite map $X\to \mathbb{A}^n$ using Noether normalization (already here it is crucial to work with constructible sheaves, not constant sheaves, so we clearly gain something from generalization),
  2. Prove the result for $\mathbb{A}^1$,
  3. Prove the result for $\mathbb{A}^n$ by induction on $n$ using the Leray spectral sequence. Here the crucial observation is that if we choose a sufficiently generic linear projection $\mathbb{A}^n\to \mathbb{A}^{n-1}$, then the stalks of the higher direct images will compute cohomology on the fibers.

So all in all it is a typical example of devissage, which I usually to dislike but slowly learn to appreciate. I think from the outline it is clear which are the key ideas, but I would still really like to see a conceptual proof.

share|improve this answer
    
Thank you! Part 3 of this argument seems to be the most difficult one; I should think about it. –  Mikhail Bondarko Oct 8 '12 at 9:11
3  
Devissage is a bit like induction, which can also be confusing. Most of the work goes into the induction step (step 3). –  Donu Arapura Oct 8 '12 at 11:12
1  
The other thing that I should point out is that for this devissage argument to work, one has to prove it for general coefficients. –  Donu Arapura Oct 8 '12 at 11:42
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.